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franse6cf5df2008-08-15 23:14:31 +02001Introduction
2============
3
4Having looked at the linux mtd/nand driver and more specific at nand_ecc.c
5I felt there was room for optimisation. I bashed the code for a few hours
6performing tricks like table lookup removing superfluous code etc.
7After that the speed was increased by 35-40%.
8Still I was not too happy as I felt there was additional room for improvement.
9
10Bad! I was hooked.
11I decided to annotate my steps in this file. Perhaps it is useful to someone
12or someone learns something from it.
13
14
15The problem
16===========
17
18NAND flash (at least SLC one) typically has sectors of 256 bytes.
19However NAND flash is not extremely reliable so some error detection
20(and sometimes correction) is needed.
21
22This is done by means of a Hamming code. I'll try to explain it in
23laymans terms (and apologies to all the pro's in the field in case I do
24not use the right terminology, my coding theory class was almost 30
25years ago, and I must admit it was not one of my favourites).
26
27As I said before the ecc calculation is performed on sectors of 256
28bytes. This is done by calculating several parity bits over the rows and
29columns. The parity used is even parity which means that the parity bit = 1
30if the data over which the parity is calculated is 1 and the parity bit = 0
31if the data over which the parity is calculated is 0. So the total
32number of bits over the data over which the parity is calculated + the
33parity bit is even. (see wikipedia if you can't follow this).
34Parity is often calculated by means of an exclusive or operation,
35sometimes also referred to as xor. In C the operator for xor is ^
36
37Back to ecc.
38Let's give a small figure:
39
40byte 0: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp4 ... rp14
41byte 1: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp2 rp4 ... rp14
42byte 2: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp4 ... rp14
43byte 3: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp4 ... rp14
44byte 4: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp5 ... rp14
45....
46byte 254: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp5 ... rp15
47byte 255: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp5 ... rp15
48 cp1 cp0 cp1 cp0 cp1 cp0 cp1 cp0
49 cp3 cp3 cp2 cp2 cp3 cp3 cp2 cp2
50 cp5 cp5 cp5 cp5 cp4 cp4 cp4 cp4
51
52This figure represents a sector of 256 bytes.
Matt LaPlante19f59462009-04-27 15:06:31 +020053cp is my abbreviation for column parity, rp for row parity.
franse6cf5df2008-08-15 23:14:31 +020054
55Let's start to explain column parity.
56cp0 is the parity that belongs to all bit0, bit2, bit4, bit6.
57so the sum of all bit0, bit2, bit4 and bit6 values + cp0 itself is even.
58Similarly cp1 is the sum of all bit1, bit3, bit5 and bit7.
59cp2 is the parity over bit0, bit1, bit4 and bit5
60cp3 is the parity over bit2, bit3, bit6 and bit7.
61cp4 is the parity over bit0, bit1, bit2 and bit3.
62cp5 is the parity over bit4, bit5, bit6 and bit7.
63Note that each of cp0 .. cp5 is exactly one bit.
64
65Row parity actually works almost the same.
66rp0 is the parity of all even bytes (0, 2, 4, 6, ... 252, 254)
67rp1 is the parity of all odd bytes (1, 3, 5, 7, ..., 253, 255)
68rp2 is the parity of all bytes 0, 1, 4, 5, 8, 9, ...
69(so handle two bytes, then skip 2 bytes).
70rp3 is covers the half rp2 does not cover (bytes 2, 3, 6, 7, 10, 11, ...)
71for rp4 the rule is cover 4 bytes, skip 4 bytes, cover 4 bytes, skip 4 etc.
72so rp4 calculates parity over bytes 0, 1, 2, 3, 8, 9, 10, 11, 16, ...)
73and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
74The story now becomes quite boring. I guess you get the idea.
75rp6 covers 8 bytes then skips 8 etc
76rp7 skips 8 bytes then covers 8 etc
77rp8 covers 16 bytes then skips 16 etc
78rp9 skips 16 bytes then covers 16 etc
79rp10 covers 32 bytes then skips 32 etc
80rp11 skips 32 bytes then covers 32 etc
81rp12 covers 64 bytes then skips 64 etc
82rp13 skips 64 bytes then covers 64 etc
83rp14 covers 128 bytes then skips 128
84rp15 skips 128 bytes then covers 128
85
86In the end the parity bits are grouped together in three bytes as
87follows:
88ECC Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
89ECC 0 rp07 rp06 rp05 rp04 rp03 rp02 rp01 rp00
90ECC 1 rp15 rp14 rp13 rp12 rp11 rp10 rp09 rp08
91ECC 2 cp5 cp4 cp3 cp2 cp1 cp0 1 1
92
93I detected after writing this that ST application note AN1823
Justin P. Mattock0ea6e612010-07-23 20:51:24 -070094(http://www.st.com/stonline/) gives a much
franse6cf5df2008-08-15 23:14:31 +020095nicer picture.(but they use line parity as term where I use row parity)
96Oh well, I'm graphically challenged, so suffer with me for a moment :-)
97And I could not reuse the ST picture anyway for copyright reasons.
98
99
100Attempt 0
101=========
102
103Implementing the parity calculation is pretty simple.
104In C pseudocode:
105for (i = 0; i < 256; i++)
106{
107 if (i & 0x01)
108 rp1 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
109 else
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800110 rp0 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp0;
franse6cf5df2008-08-15 23:14:31 +0200111 if (i & 0x02)
112 rp3 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp3;
113 else
114 rp2 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp2;
115 if (i & 0x04)
116 rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
117 else
118 rp4 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp4;
119 if (i & 0x08)
120 rp7 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp7;
121 else
122 rp6 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp6;
123 if (i & 0x10)
124 rp9 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp9;
125 else
126 rp8 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp8;
127 if (i & 0x20)
128 rp11 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp11;
129 else
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800130 rp10 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp10;
franse6cf5df2008-08-15 23:14:31 +0200131 if (i & 0x40)
132 rp13 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp13;
133 else
134 rp12 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp12;
135 if (i & 0x80)
136 rp15 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp15;
137 else
138 rp14 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp14;
139 cp0 = bit6 ^ bit4 ^ bit2 ^ bit0 ^ cp0;
140 cp1 = bit7 ^ bit5 ^ bit3 ^ bit1 ^ cp1;
141 cp2 = bit5 ^ bit4 ^ bit1 ^ bit0 ^ cp2;
142 cp3 = bit7 ^ bit6 ^ bit3 ^ bit2 ^ cp3
143 cp4 = bit3 ^ bit2 ^ bit1 ^ bit0 ^ cp4
144 cp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ cp5
145}
146
147
148Analysis 0
149==========
150
151C does have bitwise operators but not really operators to do the above
152efficiently (and most hardware has no such instructions either).
153Therefore without implementing this it was clear that the code above was
154not going to bring me a Nobel prize :-)
155
156Fortunately the exclusive or operation is commutative, so we can combine
157the values in any order. So instead of calculating all the bits
158individually, let us try to rearrange things.
159For the column parity this is easy. We can just xor the bytes and in the
160end filter out the relevant bits. This is pretty nice as it will bring
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800161all cp calculation out of the for loop.
franse6cf5df2008-08-15 23:14:31 +0200162
163Similarly we can first xor the bytes for the various rows.
164This leads to:
165
166
167Attempt 1
168=========
169
170const char parity[256] = {
171 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
172 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
173 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
174 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
175 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
176 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
177 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
178 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
179 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
180 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
181 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
182 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
183 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
184 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
185 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
186 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
187};
188
189void ecc1(const unsigned char *buf, unsigned char *code)
190{
191 int i;
192 const unsigned char *bp = buf;
193 unsigned char cur;
194 unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
195 unsigned char rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
196 unsigned char par;
197
198 par = 0;
199 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
200 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
201 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
202 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
203
204 for (i = 0; i < 256; i++)
205 {
206 cur = *bp++;
207 par ^= cur;
208 if (i & 0x01) rp1 ^= cur; else rp0 ^= cur;
209 if (i & 0x02) rp3 ^= cur; else rp2 ^= cur;
210 if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
211 if (i & 0x08) rp7 ^= cur; else rp6 ^= cur;
212 if (i & 0x10) rp9 ^= cur; else rp8 ^= cur;
213 if (i & 0x20) rp11 ^= cur; else rp10 ^= cur;
214 if (i & 0x40) rp13 ^= cur; else rp12 ^= cur;
215 if (i & 0x80) rp15 ^= cur; else rp14 ^= cur;
216 }
217 code[0] =
218 (parity[rp7] << 7) |
219 (parity[rp6] << 6) |
220 (parity[rp5] << 5) |
221 (parity[rp4] << 4) |
222 (parity[rp3] << 3) |
223 (parity[rp2] << 2) |
224 (parity[rp1] << 1) |
225 (parity[rp0]);
226 code[1] =
227 (parity[rp15] << 7) |
228 (parity[rp14] << 6) |
229 (parity[rp13] << 5) |
230 (parity[rp12] << 4) |
231 (parity[rp11] << 3) |
232 (parity[rp10] << 2) |
233 (parity[rp9] << 1) |
234 (parity[rp8]);
235 code[2] =
236 (parity[par & 0xf0] << 7) |
237 (parity[par & 0x0f] << 6) |
238 (parity[par & 0xcc] << 5) |
239 (parity[par & 0x33] << 4) |
240 (parity[par & 0xaa] << 3) |
241 (parity[par & 0x55] << 2);
242 code[0] = ~code[0];
243 code[1] = ~code[1];
244 code[2] = ~code[2];
245}
246
247Still pretty straightforward. The last three invert statements are there to
248give a checksum of 0xff 0xff 0xff for an empty flash. In an empty flash
249all data is 0xff, so the checksum then matches.
250
251I also introduced the parity lookup. I expected this to be the fastest
252way to calculate the parity, but I will investigate alternatives later
253on.
254
255
256Analysis 1
257==========
258
259The code works, but is not terribly efficient. On my system it took
260almost 4 times as much time as the linux driver code. But hey, if it was
261*that* easy this would have been done long before.
262No pain. no gain.
263
264Fortunately there is plenty of room for improvement.
265
266In step 1 we moved from bit-wise calculation to byte-wise calculation.
267However in C we can also use the unsigned long data type and virtually
268every modern microprocessor supports 32 bit operations, so why not try
269to write our code in such a way that we process data in 32 bit chunks.
270
271Of course this means some modification as the row parity is byte by
272byte. A quick analysis:
273for the column parity we use the par variable. When extending to 32 bits
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800274we can in the end easily calculate rp0 and rp1 from it.
franse6cf5df2008-08-15 23:14:31 +0200275(because par now consists of 4 bytes, contributing to rp1, rp0, rp1, rp0
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800276respectively, from MSB to LSB)
franse6cf5df2008-08-15 23:14:31 +0200277also rp2 and rp3 can be easily retrieved from par as rp3 covers the
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800278first two MSBs and rp2 covers the last two LSBs.
franse6cf5df2008-08-15 23:14:31 +0200279
280Note that of course now the loop is executed only 64 times (256/4).
281And note that care must taken wrt byte ordering. The way bytes are
282ordered in a long is machine dependent, and might affect us.
283Anyway, if there is an issue: this code is developed on x86 (to be
284precise: a DELL PC with a D920 Intel CPU)
285
286And of course the performance might depend on alignment, but I expect
287that the I/O buffers in the nand driver are aligned properly (and
288otherwise that should be fixed to get maximum performance).
289
290Let's give it a try...
291
292
293Attempt 2
294=========
295
296extern const char parity[256];
297
298void ecc2(const unsigned char *buf, unsigned char *code)
299{
300 int i;
301 const unsigned long *bp = (unsigned long *)buf;
302 unsigned long cur;
303 unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
304 unsigned long rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
305 unsigned long par;
306
307 par = 0;
308 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
309 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
310 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
311 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
312
313 for (i = 0; i < 64; i++)
314 {
315 cur = *bp++;
316 par ^= cur;
317 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
318 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
319 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
320 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
321 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
322 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
323 }
324 /*
325 we need to adapt the code generation for the fact that rp vars are now
326 long; also the column parity calculation needs to be changed.
327 we'll bring rp4 to 15 back to single byte entities by shifting and
328 xoring
329 */
330 rp4 ^= (rp4 >> 16); rp4 ^= (rp4 >> 8); rp4 &= 0xff;
331 rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
332 rp6 ^= (rp6 >> 16); rp6 ^= (rp6 >> 8); rp6 &= 0xff;
333 rp7 ^= (rp7 >> 16); rp7 ^= (rp7 >> 8); rp7 &= 0xff;
334 rp8 ^= (rp8 >> 16); rp8 ^= (rp8 >> 8); rp8 &= 0xff;
335 rp9 ^= (rp9 >> 16); rp9 ^= (rp9 >> 8); rp9 &= 0xff;
336 rp10 ^= (rp10 >> 16); rp10 ^= (rp10 >> 8); rp10 &= 0xff;
337 rp11 ^= (rp11 >> 16); rp11 ^= (rp11 >> 8); rp11 &= 0xff;
338 rp12 ^= (rp12 >> 16); rp12 ^= (rp12 >> 8); rp12 &= 0xff;
339 rp13 ^= (rp13 >> 16); rp13 ^= (rp13 >> 8); rp13 &= 0xff;
340 rp14 ^= (rp14 >> 16); rp14 ^= (rp14 >> 8); rp14 &= 0xff;
341 rp15 ^= (rp15 >> 16); rp15 ^= (rp15 >> 8); rp15 &= 0xff;
342 rp3 = (par >> 16); rp3 ^= (rp3 >> 8); rp3 &= 0xff;
343 rp2 = par & 0xffff; rp2 ^= (rp2 >> 8); rp2 &= 0xff;
344 par ^= (par >> 16);
345 rp1 = (par >> 8); rp1 &= 0xff;
346 rp0 = (par & 0xff);
347 par ^= (par >> 8); par &= 0xff;
348
349 code[0] =
350 (parity[rp7] << 7) |
351 (parity[rp6] << 6) |
352 (parity[rp5] << 5) |
353 (parity[rp4] << 4) |
354 (parity[rp3] << 3) |
355 (parity[rp2] << 2) |
356 (parity[rp1] << 1) |
357 (parity[rp0]);
358 code[1] =
359 (parity[rp15] << 7) |
360 (parity[rp14] << 6) |
361 (parity[rp13] << 5) |
362 (parity[rp12] << 4) |
363 (parity[rp11] << 3) |
364 (parity[rp10] << 2) |
365 (parity[rp9] << 1) |
366 (parity[rp8]);
367 code[2] =
368 (parity[par & 0xf0] << 7) |
369 (parity[par & 0x0f] << 6) |
370 (parity[par & 0xcc] << 5) |
371 (parity[par & 0x33] << 4) |
372 (parity[par & 0xaa] << 3) |
373 (parity[par & 0x55] << 2);
374 code[0] = ~code[0];
375 code[1] = ~code[1];
376 code[2] = ~code[2];
377}
378
379The parity array is not shown any more. Note also that for these
380examples I kinda deviated from my regular programming style by allowing
381multiple statements on a line, not using { } in then and else blocks
382with only a single statement and by using operators like ^=
383
384
385Analysis 2
386==========
387
388The code (of course) works, and hurray: we are a little bit faster than
389the linux driver code (about 15%). But wait, don't cheer too quickly.
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800390There is more to be gained.
franse6cf5df2008-08-15 23:14:31 +0200391If we look at e.g. rp14 and rp15 we see that we either xor our data with
392rp14 or with rp15. However we also have par which goes over all data.
393This means there is no need to calculate rp14 as it can be calculated from
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800394rp15 through rp14 = par ^ rp15, because par = rp14 ^ rp15;
franse6cf5df2008-08-15 23:14:31 +0200395(or if desired we can avoid calculating rp15 and calculate it from
396rp14). That is why some places refer to inverse parity.
397Of course the same thing holds for rp4/5, rp6/7, rp8/9, rp10/11 and rp12/13.
398Effectively this means we can eliminate the else clause from the if
399statements. Also we can optimise the calculation in the end a little bit
400by going from long to byte first. Actually we can even avoid the table
401lookups
402
403Attempt 3
404=========
405
406Odd replaced:
407 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
408 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
409 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
410 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
411 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
412 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
413with
414 if (i & 0x01) rp5 ^= cur;
415 if (i & 0x02) rp7 ^= cur;
416 if (i & 0x04) rp9 ^= cur;
417 if (i & 0x08) rp11 ^= cur;
418 if (i & 0x10) rp13 ^= cur;
419 if (i & 0x20) rp15 ^= cur;
420
421 and outside the loop added:
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800422 rp4 = par ^ rp5;
423 rp6 = par ^ rp7;
424 rp8 = par ^ rp9;
425 rp10 = par ^ rp11;
426 rp12 = par ^ rp13;
427 rp14 = par ^ rp15;
franse6cf5df2008-08-15 23:14:31 +0200428
429And after that the code takes about 30% more time, although the number of
430statements is reduced. This is also reflected in the assembly code.
431
432
433Analysis 3
434==========
435
436Very weird. Guess it has to do with caching or instruction parallellism
437or so. I also tried on an eeePC (Celeron, clocked at 900 Mhz). Interesting
438observation was that this one is only 30% slower (according to time)
439executing the code as my 3Ghz D920 processor.
440
441Well, it was expected not to be easy so maybe instead move to a
442different track: let's move back to the code from attempt2 and do some
443loop unrolling. This will eliminate a few if statements. I'll try
444different amounts of unrolling to see what works best.
445
446
447Attempt 4
448=========
449
450Unrolled the loop 1, 2, 3 and 4 times.
451For 4 the code starts with:
452
453 for (i = 0; i < 4; i++)
454 {
455 cur = *bp++;
456 par ^= cur;
457 rp4 ^= cur;
458 rp6 ^= cur;
459 rp8 ^= cur;
460 rp10 ^= cur;
461 if (i & 0x1) rp13 ^= cur; else rp12 ^= cur;
462 if (i & 0x2) rp15 ^= cur; else rp14 ^= cur;
463 cur = *bp++;
464 par ^= cur;
465 rp5 ^= cur;
466 rp6 ^= cur;
467 ...
468
469
470Analysis 4
471==========
472
473Unrolling once gains about 15%
474Unrolling twice keeps the gain at about 15%
475Unrolling three times gives a gain of 30% compared to attempt 2.
476Unrolling four times gives a marginal improvement compared to unrolling
477three times.
478
479I decided to proceed with a four time unrolled loop anyway. It was my gut
480feeling that in the next steps I would obtain additional gain from it.
481
482The next step was triggered by the fact that par contains the xor of all
483bytes and rp4 and rp5 each contain the xor of half of the bytes.
484So in effect par = rp4 ^ rp5. But as xor is commutative we can also say
485that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can
486eliminate rp5 (or rp4, but I already foresaw another optimisation).
487The same holds for rp6/7, rp8/9, rp10/11 rp12/13 and rp14/15.
488
489
490Attempt 5
491=========
492
493Effectively so all odd digit rp assignments in the loop were removed.
494This included the else clause of the if statements.
495Of course after the loop we need to correct things by adding code like:
496 rp5 = par ^ rp4;
497Also the initial assignments (rp5 = 0; etc) could be removed.
498Along the line I also removed the initialisation of rp0/1/2/3.
499
500
501Analysis 5
502==========
503
504Measurements showed this was a good move. The run-time roughly halved
505compared with attempt 4 with 4 times unrolled, and we only require 1/3rd
506of the processor time compared to the current code in the linux kernel.
507
508However, still I thought there was more. I didn't like all the if
509statements. Why not keep a running parity and only keep the last if
510statement. Time for yet another version!
511
512
513Attempt 6
514=========
515
516THe code within the for loop was changed to:
517
518 for (i = 0; i < 4; i++)
519 {
520 cur = *bp++; tmppar = cur; rp4 ^= cur;
521 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
522 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
523 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
524
525 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
526 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800527 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
528 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
franse6cf5df2008-08-15 23:14:31 +0200529
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800530 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; rp8 ^= cur;
franse6cf5df2008-08-15 23:14:31 +0200531 cur = *bp++; tmppar ^= cur; rp6 ^= cur; rp8 ^= cur;
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800532 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp8 ^= cur;
franse6cf5df2008-08-15 23:14:31 +0200533 cur = *bp++; tmppar ^= cur; rp8 ^= cur;
534
535 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
536 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
537 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
538 cur = *bp++; tmppar ^= cur;
539
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800540 par ^= tmppar;
franse6cf5df2008-08-15 23:14:31 +0200541 if ((i & 0x1) == 0) rp12 ^= tmppar;
542 if ((i & 0x2) == 0) rp14 ^= tmppar;
543 }
544
545As you can see tmppar is used to accumulate the parity within a for
Masanari Iidaac20ad12013-08-20 05:53:49 -0700546iteration. In the last 3 statements is added to par and, if needed,
franse6cf5df2008-08-15 23:14:31 +0200547to rp12 and rp14.
548
549While making the changes I also found that I could exploit that tmppar
550contains the running parity for this iteration. So instead of having:
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800551rp4 ^= cur; rp6 ^= cur;
552I removed the rp6 ^= cur; statement and did rp6 ^= tmppar; on next
franse6cf5df2008-08-15 23:14:31 +0200553statement. A similar change was done for rp8 and rp10
554
555
556Analysis 6
557==========
558
559Measuring this code again showed big gain. When executing the original
560linux code 1 million times, this took about 1 second on my system.
561(using time to measure the performance). After this iteration I was back
562to 0.075 sec. Actually I had to decide to start measuring over 10
Matt LaPlante19f59462009-04-27 15:06:31 +0200563million iterations in order not to lose too much accuracy. This one
franse6cf5df2008-08-15 23:14:31 +0200564definitely seemed to be the jackpot!
565
566There is a little bit more room for improvement though. There are three
567places with statements:
568rp4 ^= cur; rp6 ^= cur;
569It seems more efficient to also maintain a variable rp4_6 in the while
570loop; This eliminates 3 statements per loop. Of course after the loop we
571need to correct by adding:
572 rp4 ^= rp4_6;
573 rp6 ^= rp4_6
Matt LaPlante19f59462009-04-27 15:06:31 +0200574Furthermore there are 4 sequential assignments to rp8. This can be
575encoded slightly more efficiently by saving tmppar before those 4 lines
franse6cf5df2008-08-15 23:14:31 +0200576and later do rp8 = rp8 ^ tmppar ^ notrp8;
577(where notrp8 is the value of rp8 before those 4 lines).
578Again a use of the commutative property of xor.
579Time for a new test!
580
581
582Attempt 7
583=========
584
585The new code now looks like:
586
587 for (i = 0; i < 4; i++)
588 {
589 cur = *bp++; tmppar = cur; rp4 ^= cur;
590 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
591 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
592 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
593
594 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
595 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800596 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
597 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
franse6cf5df2008-08-15 23:14:31 +0200598
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800599 notrp8 = tmppar;
600 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
franse6cf5df2008-08-15 23:14:31 +0200601 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800602 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
franse6cf5df2008-08-15 23:14:31 +0200603 cur = *bp++; tmppar ^= cur;
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800604 rp8 = rp8 ^ tmppar ^ notrp8;
franse6cf5df2008-08-15 23:14:31 +0200605
606 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
607 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
608 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
609 cur = *bp++; tmppar ^= cur;
610
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800611 par ^= tmppar;
franse6cf5df2008-08-15 23:14:31 +0200612 if ((i & 0x1) == 0) rp12 ^= tmppar;
613 if ((i & 0x2) == 0) rp14 ^= tmppar;
614 }
615 rp4 ^= rp4_6;
616 rp6 ^= rp4_6;
617
618
619Not a big change, but every penny counts :-)
620
621
622Analysis 7
623==========
624
Matt LaPlante19f59462009-04-27 15:06:31 +0200625Actually this made things worse. Not very much, but I don't want to move
franse6cf5df2008-08-15 23:14:31 +0200626into the wrong direction. Maybe something to investigate later. Could
627have to do with caching again.
628
629Guess that is what there is to win within the loop. Maybe unrolling one
630more time will help. I'll keep the optimisations from 7 for now.
631
632
633Attempt 8
634=========
635
636Unrolled the loop one more time.
637
638
639Analysis 8
640==========
641
642This makes things worse. Let's stick with attempt 6 and continue from there.
643Although it seems that the code within the loop cannot be optimised
644further there is still room to optimize the generation of the ecc codes.
Matt LaPlante19f59462009-04-27 15:06:31 +0200645We can simply calculate the total parity. If this is 0 then rp4 = rp5
franse6cf5df2008-08-15 23:14:31 +0200646etc. If the parity is 1, then rp4 = !rp5;
647But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits
648in the result byte and then do something like
649 code[0] |= (code[0] << 1);
650Lets test this.
651
652
653Attempt 9
654=========
655
656Changed the code but again this slightly degrades performance. Tried all
657kind of other things, like having dedicated parity arrays to avoid the
658shift after parity[rp7] << 7; No gain.
659Change the lookup using the parity array by using shift operators (e.g.
660replace parity[rp7] << 7 with:
661rp7 ^= (rp7 << 4);
662rp7 ^= (rp7 << 2);
663rp7 ^= (rp7 << 1);
664rp7 &= 0x80;
665No gain.
666
667The only marginal change was inverting the parity bits, so we can remove
668the last three invert statements.
669
670Ah well, pity this does not deliver more. Then again 10 million
671iterations using the linux driver code takes between 13 and 13.5
672seconds, whereas my code now takes about 0.73 seconds for those 10
673million iterations. So basically I've improved the performance by a
674factor 18 on my system. Not that bad. Of course on different hardware
675you will get different results. No warranties!
676
677But of course there is no such thing as a free lunch. The codesize almost
678tripled (from 562 bytes to 1434 bytes). Then again, it is not that much.
679
680
681Correcting errors
682=================
683
684For correcting errors I again used the ST application note as a starter,
685but I also peeked at the existing code.
686The algorithm itself is pretty straightforward. Just xor the given and
687the calculated ecc. If all bytes are 0 there is no problem. If 11 bits
688are 1 we have one correctable bit error. If there is 1 bit 1, we have an
689error in the given ecc code.
690It proved to be fastest to do some table lookups. Performance gain
691introduced by this is about a factor 2 on my system when a repair had to
692be done, and 1% or so if no repair had to be done.
693Code size increased from 330 bytes to 686 bytes for this function.
694(gcc 4.2, -O3)
695
696
697Conclusion
698==========
699
700The gain when calculating the ecc is tremendous. Om my development hardware
701a speedup of a factor of 18 for ecc calculation was achieved. On a test on an
702embedded system with a MIPS core a factor 7 was obtained.
Wang YanQingfc5adbe2015-10-30 00:36:33 +0800703On a test with a Linksys NSLU2 (ARMv5TE processor) the speedup was a factor
franse6cf5df2008-08-15 23:14:31 +02007045 (big endian mode, gcc 4.1.2, -O3)
705For correction not much gain could be obtained (as bitflips are rare). Then
706again there are also much less cycles spent there.
707
708It seems there is not much more gain possible in this, at least when
709programmed in C. Of course it might be possible to squeeze something more
710out of it with an assembler program, but due to pipeline behaviour etc
711this is very tricky (at least for intel hw).
712
713Author: Frans Meulenbroeks
714Copyright (C) 2008 Koninklijke Philips Electronics NV.